Question

A spherical ball A of surface area 20 $c{m}^2$ is kept at the centre of a hollow spherical shell B of area 80 $c{m}^2$. The surface of A and the inner surface of B emit as blackbodies. Assume that the thermal conductivity of the material of B is extremely poor and that of A is very high and that the air between A and B has been pumped out. The heat capacities of A and B are 42 J/${}^{\circ }$C and 82 J/ ${}^{\circ }$ C respectively. Initially the temperature of A is 100 ${}^{\circ }$ C and that of B is 20 ${}^{\circ }$ C.Find the rate of change of temperature of A and that of B at this instant.

• A. ${0.005}^{\circ }c/sand{0.0001}^{\circ }c/s$
• B. ${0.05}^{\circ }c/sand{0.01}^{\circ }c/s$
• C. ${0.025}^{\circ }c/sand{0.0050}^{\circ }c/s$
• D. ${0.03}^{\circ }c/sand{0.01}^{\circ }c/s$

Answer 1

The correct option is D

${0.03}^{\circ }c/sand{0.01}^{\circ }c/s$

The inner surface of B radiates at a rate of $\frac{\Delta Q}{\Delta t}=\sigma A_B{T}_A^4$,

where $\sigma \to$ stefan's constant

AB $\to$Area

TB $\to$ Temperature

Similarly, A radiates $\frac{\Delta Q_2}{\Delta t}=\sigma A_B{T}_A^4$

Now, all the radiation from inner surface of B goes inside the hollow of the shell, as B is a very poor conductor. So, this radiation falls on A, and A being a blackbody absorbs all the radiation falling on it! And it being a good conductor the heat travels to the centre almost instantly, thus making the temperature uniform throughout the ball. Similarly, all the heat radiated by A is completely absorbed by inner surface of B, and none of it is lost to the surrounding as B is a very poor conductor.

∴ Net heat lost by B = Net heat gained by A =$\sigma (A_B{T}_B^4-A_A{T}_A^4)$

= $5.67\times {10}^{-8}\times (80\times 293.1s^4-20\times 373.1s^4)\times {10}^{-4}$

= 1.15 w

Now this is equal to specific heat $\times \frac{dtemperature}{dTime}$ of both A and B

$\therefore \frac{d{T}_A}{dt}={\frac{1.15}{42}}^{\circ }c/s={0.03}^{\circ }c/s$

$\frac{d{T}_B}{dt}={\frac{1.15}{82}}^{\circ }c/s={0.01}^{\circ }c/s$